Finding the partial sums of an infinite series I've tried to find the partial sums of this infinite series in order to get the series sum result:
$$\sum_{n=1}^\infty \sin\left(\frac{1}{2^n}\right )\cos\left (\frac{1}{2^n}\right )$$
Can anyone show me the steps please .. thanks
 A: Disclaimer: This is only meant to approximate the solution.
Notice that
$$2\sin x\cos x=\sin2x$$
So your sum can be reduced to
$$\frac12\sum_{n=1}^\infty\sin(2^{1-n})=\frac12\sum_{n=0}^\infty\sin(2^{-n})$$
Notice then that for all $n\ge0$ that $\sin(2^{-n})<2^{-n}$ so that
$$\begin{align}\sum_{n=0}^\infty\sin(2^{-n})&=\sum_{n=0}^{k-1}\sin(2^{-n})+\sum_{n=k}^\infty\sin(2^{-n})\\&<\sum_{n=0}^{k-1}\sin(2^{-n})+\sum_{n=k}^\infty2^{-n}\\&=\sum_{n=0}^{k-1}\sin(2^{-n})+2^{1-k}\end{align}$$
Thus, you can quickly approximate the solution.
$$\begin{align}\sum_{n=0}^\infty\sin(2^{-n})&<\sin(1)+1\approx1.841470985\\\sum_{n=0}^\infty\sin(2^{-n})&<\sin(1)+\sin(1/2)+1/2\approx1.820896523\\\sum_{n=0}^\infty\sin(2^{-n})&<\sin(1)+\sin(1/2)+\sin(1/4)+1/4\approx1.818300483\end{align}$$
If one takes this all the way, to say, $k=10$,
$$\sum_{n=0}^\infty\sin(2^{-n})\approx1.817928721$$
Thus, half of this is
$$\sum_{n=1}^\infty \sin\left(\frac{1}{2^n}\right )\cos\left (\frac{1}{2^n}\right )\approx0.908964361$$
